Question

The enthalpy change $\left( {\Delta H} \right)$ for the reaction, ${N_2}\left( g \right) + 3{H_2}\left( g \right) \to 2N{H_3}\left( g \right)$ is -92.38kJ at 298 K. The internal energy change $\Delta U$ at 298 K is:
A. -92.38kJ
B. -87.42kJ
C. -97.34kJ
D. -89.9kJ

Answer 1

Hint: Ionic character percentage of a molecule can be obtained by using the formula $\dfrac{{Actual\,dipole\,moment}}{{Calculated\,dipole\,moment}} \times 100$. The actual dipole moment is already given in the question, so obtain the calculated dipole moment using the formula $\mu = BL \times q$, where BL is the bond length and q is the charge of the electron. Bond length is the length of interatomic spacing.

Complete step by step answer:
We are given that the enthalpy change $\left( {\Delta H} \right)$ for the reaction, ${N_2}\left( g \right) + 3{H_2}\left( g \right) \to 2N{H_3}\left( g \right)$ is -92.38kJ at 298 K.

Enthalpy is the transfer of energy in a reaction. Enthalpy change is the total amount of energy released or absorbed while transforming reactants into products at some temperature and constant pressure.

Enthalpy change is represented as $\Delta H$ and can be calculated using the below formula
$\Delta H = \Delta U + \Delta nRT$, where $\Delta U$ is the internal energy change, T is the temperature in Kelvin, $\Delta n$ is the difference of no. of moles of products and reactants and R is the ideal gas constant.

In the reaction ${N_2}\left( g \right) + 3{H_2}\left( g \right) \to 2N{H_3}\left( g \right)$, the no. of moles present at products side is 2 and the no. of moles present at reactants side is 4.
Therefore,
$
\Rightarrow \Delta n = {n_p} - {n_r} \\
\Rightarrow {n_p} = 2,{n_r} = 4 \\
\Rightarrow \Delta n = 2 - 4 = - 2mol \\
 $
The temperature given is 298 K, The value of R is $8.31 \times {10^{ - 3}}kJ.{K^{ - 1}}.mo{l^{ - 1}}$
By substituting all the values in the enthalpy change equation, we get
$
\Rightarrow \Delta H = \Delta U + \Delta nRT \\
\Rightarrow \Delta U = \Delta H - \Delta nRT \\
\Rightarrow \Delta H = - 92.38kJ,T = 298K,R = 8.31 \times {10^{ - 3}}kJ.{K^{ - 1}}.mo{l^{ - 1}},\Delta n = - 2mol,\Delta U = ? \\
\Rightarrow \Delta U = - 92.38 - \left[ {\left( { - 2} \right) \times 8.31 \times {{10}^{ - 3}} \times 298} \right] \\
\Rightarrow \Delta U = - 92.38 - \left( { - 4.95} \right) \\
\Rightarrow \Delta U = - 92.38 + 4.95 \\
  \therefore \Delta U = - 87.42kJ \\
 $
Therefore, the value of internal energy change is -87.42kJ.

Hence we can conclude that option B is correct.

Note: Do not confuse enthalpy change with internal energy change, because the internal energy includes the energy of molecular structure like translational energy and enthalpy is the sum of internal energy and the energy due to the volume. Enthalpy change can be calculated only at constant pressures. The internal energy change is equal to the enthalpy change only when the volume is constant and the no. of moles are the same on both products and reactants side.